The question is in the title.
So far I have written it as
∫x²-(x-1)²dx
and I am struggling with integrating the x^-1 as I don't know if it would be 2lnx or something else.
I've done the first x
∫x² dx =x³/3

(Original post by TheGreatPumpkin)
The question is in the title.
So far I have written it as
∫x²-(x-1)²dx
and I am struggling with integrating the x^-1 as I don't know if it would be 2lnx or something else.
I've done the first x
∫x² dx =x³/3

Expand

The result of is , that is the first term squared, plus two times the product, plus the last term squared.

(Original post by TheGreatPumpkin)
The question is in the title.
So far I have written it as
∫x²-(x-1)²dx
and I am struggling with integrating the x^-1 as I don't know if it would be 2lnx or something else.
I've done the first x
∫x² dx =x³/3

You can't go from ∫(x-1∕x)² to ∫x^2 - (x^-1)^2. You need to use the reverse chain rule so you first need to differentiate the bracket so you get dy/dx = 1 + x^-2. Using the reverse chain rule you get ((x-1/x)^3)/(3(1+x^-2)). Any math geniuses agree? Quite new to integration so sorry for any mistake. Hope it helps

(Original post by samendrag)
You can't go from ∫(x-1∕x)² to ∫x^2 - (x^-1)^2. You need to use the reverse chain rule so you first need to differentiate the bracket so you get dy/dx = 1 + x^-2. Using the reverse chain rule you get ((x-1/x)^3)/(3(1+x^-2)). Any math geniuses agree? Quite new to integration so sorry for any mistake. Hope it helps